str_extract_all：返回在字符串中找到的所有组合为向量的模式 [英] str_extract_all: return all patterns found in string concatenated as vector

问题描述

我想提取除模式以外的所有内容，并将其转化为字符串。

I want to extract everything but a pattern and return this concetenated in a string.

我试图将str_extract_all与sapply和cat结合在一起

I tried to combine str_extract_all together with sapply and cat

x = c("a_1","a_20","a_40","a_30","a_28")
data <- tibble(age = x)


# extracting just the first pattern is easy
data %>% 
  mutate(age_new = str_extract(age,"[^a_]"))
# combining str_extract_all and sapply doesnt work
data %>% 
  mutate(age_new = sapply(str_extract_all(x,"[^a_]"),function(x) cat(x,sep="")))


class(str_extract_all(x,"[^a_]"))
sapply(str_extract_all(x,"[^a_]"),function(x) cat(x,sep=""))

返回NULL代替串联模式

Returns NULL instead of concatenated patterns

推荐答案

代替 cat ，我们可以使用粘贴。另外，使用 tidyverse ，可以使用 map 和 str_c （代替粘贴-来自 stringr ）

Instead of cat, we can use paste. Also, with tidyverse, can make use of map and str_c (in place of paste - from stringr)

library(tidyverse)
data %>% 
  mutate(age_new = map_chr(str_extract_all(x, "[^a_]+"), ~ str_c(.x, collapse="")))

---

使用OP的代码

---

using `OP's code

data %>%
    mutate(age_new = sapply(str_extract_all(x,"[^a_]"),
               function(x) paste(x,collapse="")))

< hr>

如果打算获取数字

---

If the intention is to get the numbers

library(readr)
data %>%
     mutate(age_new = parse_number(x))

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